0
$\begingroup$

Let $X$ be a connected locally path connected space. I want to show that it is also path connected.

Following a suggestion in this answer, fix $a\in X$ and consider the set

$$U_a = \{x\in X : \text{there is a path from } x \text{ to } a\}$$

and show that it is clopen.

If the set is clopen, then we can decompose $X$ as a disjoint union of the open sets $U_a$ and $X\setminus U_a$. Since $X$ is connected and $U_a \not= \varnothing$ ($a\in U_a$), it follows that $U_a = X$.

My question concerns how to show that $U_a$ is clopen. I considered writing $U_a$ and $X\setminus U_a$ as unions of open sets as follows:

$$U_a = \bigcup \{\text{path-connected open neighbourhoods of $a$}\}$$

$$X\setminus U_a = \bigcup_{x\in X}\bigcup\{\text{path-connected open neighbourhoods of $x$ without $a$}\}$$

It is clear that any path-connected neighbourhood of $a$ is a subset of $U_a$, but if there is a path from $x$ to $a$, does it follow that $x$ is in a path connected neighbourhood of a?

$\endgroup$
  • 1
    $\begingroup$ I think it is easier to show that every element in $U_a$ has a neighbourhood that is contained in $U_a$, and the same thing for the complement. $\endgroup$ – asdq Apr 30 at 17:09
0
$\begingroup$

Thanks to @asdq ’s suggestion, I can complete the proof as follows:

Any $x \in U_a$ has a path-connected neighborhood, say $V$. Any point in $V$ can be connected by a path to $a$ by first connecting it to $x$, and concatenating with the path to $a$. Hence, $V \subset U_a$, so $U_a$ is open.

If $x \notin U_a$, then there is a path-connected neighborhood $W$ of $x$ such that no point of $W$ connects to $a$ via a path; otherwise we could join that path to $x$, so $x \in U_a$. Therefore, $W \subset X\setminus U_a$, so $X\setminus U_a$ is open.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.